G-invariant. Define f:A [ R by f(?r) = Tr(*(Y')). Then f is
everywhere finite, lower semicontinuous, and not identically
zero. Let U = {*€A : Tr(*(y') )£a0). Then U is a Baire space, so
the function f has a point of continuity in U for suitable a, say
at K A. So f is continuous and non-zero at K .
o o
Let X be the largest eigenvalue of K (y'), and let
*:I R •[CI] be a weakly increasing continuous function with
0(t) = 1 for X^t and with 0(t) = 0 for t^w, where u0 is greater
than or equal to the second-largest eigenvalue of *Q(y')• Using
the functional calculus, set y = 0(y')- Then ||y| | = 1, y is still
G-invariant, and K (y) is a finite-rank projection. Furthermore,
*(y) is of finite rank for all *€A (since Tr(y') can have only
finitely many eigenvalues greater than u, and each occurs with
finite multiplicity), and the map K ~ Tr(ir(y)) is also continuous
at *Q, by [Di, 4.4.2(i)].
To proceed further, note that since G is compact and A is
1* , the orbit Gx is closed in A by [MR, Lemma 4.1], hence
corresponds to a G-quotient B of A with G-spectrum homeomorphic
to G/H, where H is the stabilizer group of ^ in G, via the
implication (1)=(6) in [Gl, Theorem 1]. Since *0(Y) *S a
finite-rank projection and y is G-invariant, y maps to a
G-invariant projection y in B.
To continue in this fashion for arbitrary G is difficult, so
assume now that G = T is a one-dimensional torus. Then by Theorem
2.5, B$K is isomorphic to C(G/H)$/f, and the action a of G on BQK
is exterior equivalent to the translation action B, where £.f(g)
fft"1^), for f€C(G/H,Jf). We claim that y dominates a
G-invariant rank-one projection p in B. To see this is trivial if
G = H, so we may assume H is finite cyclic. We may as well assume
that A (hence B) is stable, so that the action a of G on
B S C(G/H,K) is given by
atf(g) = vt(g)f(t"1g)vt(g)* (t€G, g€G/H)
for some cocycle
V:G C(G/Hf!/) ,
where V is the infinite-dimensional unitary group with the strong
Previous Page Next Page